Question

Two resistance when connected in parallel give
resultant value of 2ohm when connected in series the
value becomes 9ohm calculate the value of each
​

Answer 1

Given :-

• Two resistances when connected in parallel give resultant of 2Ω .
• When they are connected in series it gives resistance of 9Ω .

To Find :-

• The value of each resistance .

Answer :-

Let us take :

• $\sf \blue{First\: resistance\:be\:x.}$
• $\sf \blue{Second\: resistance\:be\:y.}$

$\tt Now\:we\:know\:when\::-$

$\sf{\pink{\blue{\mapsto}When\: resistances\:are\: connected\:in\: series:}}$

$\boxed{\red{\bf{\pink{\dag}\:R_{net}=R_1+R_2+R_3+......+R_n}}}$

$\bf\qquad\qquad\qquad And,$

$\sf{\pink{\blue{\mapsto}When\: resistances\:are\: connected\:in\: parallel:}}$

$\boxed{\red{\bf{\pink{\dag}\:\dfrac{1}{R_{net}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+......+\dfrac{1}{R_n}}}}$

$\tt Let's\:use\:these\:one\:by\:one,$

$\tt{\orange{\mapsto}\green{Case\:1:In\: series}}$

$\tt:\implies R_{net}=R_1+R_2+R_3+......+R_n$

$\tt:\implies R_{net}=x+y$

$\tt:\implies 9\Omega=x+y$

$\underline{\boxed{\red{\tt\longmapsto x+y=9\Omega}}}$

___________________________________

$\tt{\orange{\mapsto}\green{Case\:2:In\: series}}$

$\tt:\implies \dfrac{1}{R_{net}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+......+\dfrac{1}{R_n}$

$\tt:\implies \dfrac{1}{R_{net}}=\dfrac{1}{x}+\dfrac{1}{y}$

$\tt:\implies \dfrac{1}{R_{net}}=\dfrac{x+y}{xy}$

$\tt:\implies \dfrac{1}{2\Omega} =\dfrac{x+y}{xy}$

$\tt:\implies 2(x+y)=xy$

$\tt:\implies 2(x+9-x)=x(9-x)$

$\tt:\implies 2\times9=9x-x^2$

$\tt:\implies 18=9x-x^2$

$\tt:\implies x^2-9x+18=0$

$\tt:\implies x^2-3x-6x+18=0$

$\tt:\implies x(x-3)-6(x-3)=0$

$\tt:\implies(x-6)(x-3)=0$

$\underline{\boxed{\red{\tt\longmapsto x\qquad=\qquad 3\Omega,6\Omega}}}$

$\boxed{\green{\bf\pink{\dag}\:Hence\:two\: resistances\:are\:6\Omega\&\:3\Omega.}}$

Answer 2

let the each resistance be R¹ and R²

in parallel

1/R¹+1/R² = R¹+R²/R¹R² =2( given )...........2

in series

R¹+R² = 9 (given )...........1

put eq1 in eq2

9/R¹R² = 2

R¹R² = 2/9

formula (a+b)²-(a-b)² = 4ab

(R¹+R²) ² + (R¹-R²)² = 4R¹R²

9² + (R¹-R²)² = 8/9

(R¹-R²)² = 8/9 - 81 = 8 - 729= 721

(R¹-R²) = 26.8

2R¹ = 35.8

R¹ = 35.8/2 = 17.9

put R¹ in any eq

R² = 8.9

hence the R¹ = 17.9 and R² = 8.9